In the reaction M^{+x} + MnO_4^- to MO_3^- + | vedclass

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(C) The given reaction is $M^{+x} + MnO_4^- 	o MO_3^- + Mn^{+2}$.Given that $1.67 \ mol$ of $M^{+x}$ reacts with $1 \ mol$ of $MnO_4^-$. Since $1.67

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$2$

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(C) The given reaction is $M^{+x} + MnO_4^- 	o MO_3^- + Mn^{+2}$.Given that $1.67 \ mol$ of $M^{+x}$ reacts with $1 \ mol$ of $MnO_4^-$. Since $1.67 \approx 5/3$,we can write the stoichiometry as:$\frac{5}{3} M^{+x} + 1 MnO_4^- 	o \frac{5}{3} MO_3^- + 1 Mn^{+2}$.Multiplying by $3$ to clear the fraction:$5 M^{+x} + 3 MnO_4^- 	o 5 MO_3^- + 3 Mn^{+2}$.Now,calculate the change in oxidation states:For $M$: $5 	imes (5 - x) = 25 - 5x$ (increase in oxidation state).For $Mn$: $3 	imes (7 - 2) = 3 	imes 5 = 15$ (decrease in oxidation state).For a balanced redox reaction,the total increase in oxidation state must equal the total decrease:$25 - 5x = 15$.$5x = 10$.$x = 2$.

In the reaction $M^{+x} + MnO_4^- \to MO_3^- + Mn^{+2}$,if $1 \ mol$ of $MnO_4^-$ oxidizes $1.67 \ mol$ of $M^{+x}$ to $MO_3^-$,find the value of $x$.

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